Answer
$\frac{1}{x} - \frac{2}{x+3} + \frac{1}{x-1}$
Work Step by Step
$\frac{7x-3}{(x^{3} + 2x^{2} - 3x)}$ = $\frac{A}{x} + \frac{B}{x+3} + \frac{C}{x-1}$
Multiply both sides by (x) (x+3) (x-1)
$7x-3 = A(x+3) (x-1) + B (x) (x-1) + C (x) (x+3)$
Substitute x=0
-3A = -3
A = 1
Substitute x=1
4C = 4
C =1
Substitute x= -3
12B = -24
B = -2
$\frac{1}{x} - \frac{2}{x+3} + \frac{1}{x-1}$
